Let \(f\) be a function from \(\mathbb{R}\) to \(\mathbb{R}\) and let \(a \in \mathbb{R}\). From the definition of the derivative, prove that if \(f\) is differentiable at \(a\) then \(f\) is continuous at \(a\).
Prove the product rule, \((fg)' = f'g + g'f\).
Solution
Goal: Prove that differentiability implies continuity at a point, and deduce the product rule for derivatives using limit laws.
1. Part (i): Differentiability at \(a\) implies continuity at \(a\).Proof:
1.1. By definition of differentiability, the limit\[f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a}\]exists as a real number.
1.2. For all \(x \neq a\), we have the algebraic identity:\[f(x) - f(a) = \frac{f(x) - f(a)}{x - a} \cdot (x - a).\]
1.3. Applying the product law for limits:\[\lim_{x \to a} (f(x) - f(a)) = \left(\lim_{x \to a} \frac{f(x) - f(a)}{x - a}\right) \cdot \left(\lim_{x \to a} (x - a)\right) = f'(a) \cdot 0 = 0.\]
1.4. Therefore \(\lim_{x \to a} f(x) = f(a)\), which proves that \(f\) is continuous at \(a\).
2. Part (ii): Product rule \((fg)'(a) = f'(a)g(a) + f(a)g'(a)\).Proof:
2.1. Suppose \(f\) and \(g\) are differentiable at \(a\). The difference quotient for the product \(fg\) is:\[\frac{(fg)(x) - (fg)(a)}{x - a} = \frac{f(x)g(x) - f(a)g(a)}{x - a}.\]
2.2. Add and subtract the cross-term \(f(a)g(x)\) in the numerator:\[\frac{f(x)g(x) - f(a)g(a)}{x - a} = \frac{[f(x)g(x) - f(a)g(x)] + [f(a)g(x) - f(a)g(a)]}{x - a} = \frac{f(x) - f(a)}{x - a} g(x) + f(a) \frac{g(x) - g(a)}{x - a}.\]
2.3. Since \(g\) is differentiable at \(a\), by 1 \(g\) is continuous at \(a\), so \(\lim_{x \to a} g(x) = g(a)\).
2.4. Applying limit laws for sums and products as \(x \to a\):\[\begin{aligned} (fg)'(a) &= \lim_{x \to a} \left[ \frac{f(x) - f(a)}{x - a} g(x) + f(a) \frac{g(x) - g(a)}{x - a} \right] \\ &= \left(\lim_{x \to a} \frac{f(x) - f(a)}{x - a}\right) \cdot \left(\lim_{x \to a} g(x)\right) + f(a) \cdot \left(\lim_{x \to a} \frac{g(x) - g(a)}{x - a}\right) \\ &= f'(a) g(a) + f(a) g'(a). \end{aligned}\]
2.5. Thus \((fg)' = f'g + fg'\). Q.E.D.